# Set of 2-Dimensional Real Orthogonal Matrices is Compact in Normed Real Square Matrix Vector Space

## Theorem

Let $\struct {\R^{2 \times2}, \norm {\, \cdot \,}_\infty}$ be the normed 2-dimensional real square matrix vector space.

Let $\map O 2 := \set {\mathbf R \in \R^{2 \times2}: \mathbf R^\intercal \mathbf R = \mathbf I_2}$ be the orthogonal group of degree $2$ over real numbers.

Then $\map O 2$ is a compact set in $\struct {\R^{2 \times 2}, \norm {\, \cdot \,}_\infty}$.

## Proof

Let $\sequence {\mathbf R_n}_{n \mathop \in \N}$ be a sequence in $\map O 2$.

Let:

$\begin{bmatrix} a_n & b_n \\ c_n & d_n \\ \end{bmatrix} := \mathbf R_n$

where $\sequence {a_n}_{n \mathop \in \N}$, $\sequence {b_n}_{n \mathop \in \N}$, $\sequence {c_n}_{n \mathop \in \N}$, $\sequence {d_n}_{n \mathop \in \N} \in \R$ are real sequences.

By definition of $\mathbf R_n$:

$\begin{bmatrix} a_n & c_n \\ b_n & d_n \\ \end{bmatrix} \begin{bmatrix} a_n & b_n \\ c_n & d_n \\ \end{bmatrix} = \begin{bmatrix} a_n^2 + c_n^2 & a_n b_n + c_n d_n \\ a_n b_n + c_n d_n & b_n^2 + d_n^2 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$

We have that:

$a_n^2 + c_n^2 = 1$
$b_n^2 + d_n^2 = 1$

Hence:

$-1 \le a_n, b_n, c_n, d_n \le 1$

By Bolzano-Weierstrass theorem we can find subsequences $\sequence {a_{n_k} }_{k \mathop \in \N}$, $\sequence {b_{n_k} }_{k \mathop \in \N}$, $\sequence {c_{n_k} }_{k \mathop \in \N}$, $\sequence {d_{n_k} }_{k \mathop \in \N}$ such that:

$\ds \lim_{k \to \infty} a_{n_k} = a$
$\ds \lim_{k \to \infty} b_{n_k} = b$
$\ds \lim_{k \to \infty} c_{n_k} = c$
$\ds \lim_{k \to \infty} d_{n_k} = d$

In other words, given $\epsilon \in \R_{>0}$:

$\exists N_a \in \N: \forall n \in \N: n > N_a \implies \size {a_n - a} < \epsilon$
$\exists N_b \in \N: \forall n \in \N: n > N_b \implies \size {b_n - b} < \epsilon$
$\exists N_c \in \N: \forall n \in \N: n > N_c \implies \size {c_n - c} < \epsilon$
$\exists N_d \in \N: \forall n \in \N: n > N_d \implies \size{d_n - d} < \epsilon$

Let $N := \max \set {N_a, N_b, N_c, N_d}$

Then:

$\exists N \in \N: \forall n \in \N: n > N \implies \paren {\size {a_n - a} < \epsilon} \land \paren {\size {b_n - b} < \epsilon} \land \paren {\size {c_n - c} < \epsilon} \land \paren {\size {d_n - d} < \epsilon}$

The set of $\size {a_n - a}, \size {b_n - b}, \size {c_n - c}, \size {d_n - d}$ constitutes a finite subset of the set of real numbers which is ordered.

Without loss of generality, let it be $\size {a_n - a}$.

Then:

$\ds \exists N \in \N: \forall n \in \N: n > N \implies \size {a_n - a} = \max_{\begin {split} & 1 \mathop \le i \mathop \le 2\\ & 1 \mathop \le j \mathop \le 2 \end {split} } \size { {r_n}_{~i j} - r_{i j} } = \norm {\mathbf R_n - \mathbf R}_\infty < \epsilon$

where:

$\mathbf R = \sqbrk r_{i j}$ and $\mathbf R_n = \sqbrk {r_n}_{i j}$

We have that $\mathbf R_n^\intercal \mathbf R_n = \mathbf I_2$.

By combination theorem for real sequences, it follows that $\mathbf R^\intercal \mathbf R = \mathbf I_2$.

Hence, $\mathbf R \in \map O 2$.

By definition, $\map O 2$ is compact.

$\blacksquare$